For a ring R with an ideal I, the I-adic topology makes R into a topological ring Let $R$ be a commutative ring with identity. Let $I$ be an ideal of $R$. Suppose, we give a topology on $R$ where a set is open if and only if it is a union of cosets of powers of $I$. Then, is $R$ a topological ring?
EDIT: The question has been edited in the light of comments below.
 A: That cannot work: the topology you want has as a basis the set of all translations of all powers of I, while not all sets you describe in the first paragraph are unions of such things.
For example, suppose $I$ is such that $I^2=0$. Then the open sets are just the cosets of $I$ in $R$. But there are sets which contain a coset but are not a union of cosets. We can construct a concrete example as follows: Let $k$ be a field, and let $R=k[x]/(x^2)$. Let $\varepsilon$ be the image of $x\in k[x]$ in the quotient $R$. Then $I=(\varepsilon)$ is an ideal which squares to zero. And $I\cup(k\setminus\{0\})$ is a set which contains a power of $I$, but it is not a union of cosets of $I$.
A: In another, now deleted, venue you mentioned a more specific question of just showing the operations are continuous.  You mentioned you could handle negation, but had trouble with addition and multiplication.  Here is one way to verify them:
Let J = In.
The inverse image of x+J under addition is { (y,z) : y+z-x in J }. This is a union of A(y) = (y+J)⊕(x-y+J) ≤ R⊕R as y varies over R. Each A(y) is open (being a direct product of open sets), and so the union is open. In other words, the preimage under + of an open set is open.
The inverse image x+J under multiplication is { (y,z) : yz-x in J }. This is a union of B(y) = (y+J)⊕(∪{ z+J : yz-x in J }). Each B(y) is open (being a direct product of an open set and a union of open sets). In other words the preimage under multiplication of an open set is open.
The only trick to it is (1) knowing some open sets of a direct product and (2) noticing that you could have just worked in a quotient ring R/J where every set (every!) is open, since it is a union of singletons, x+J. In plainer language, (y+j)*(z+j') = yz + jz+yj'+jj' = yz + (something in J) is still in yz+J.
